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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 26

  • question_answer
    The equilibrium, \[{{P}_{4(g)}}+6C{{l}_{2(g)}}4PC{{l}_{3(g)}}\]is attained by mixing equal moles of \[{{P}_{4}}\]and \[C{{l}_{2}}\]in an evacuated vessel. Then at equilibrium

    A) \[[C{{l}_{2}}]>[PC{{l}_{3}}]\]

    B) \[[C{{l}_{2}}]>[{{P}_{4}}]\]

    C) \[[{{P}_{4}}]>[C{{l}_{2}}]\]   

    D) \[[PC{{l}_{3}}]>[{{P}_{4}}]\]

    Correct Answer: C

    Solution :

    [c]:\[{{P}_{4(g)}}+6C{{l}_{2(g)}}4PC{{l}_{3(g)}}\] One mole of \[{{P}_{4}}\]reacts with 6 moles of \[C{{l}_{2}}\] i.e., at equilibrium \[C{{l}_{2}}\]is consumed more than \[{{P}_{4}}\]. If we start the reaction with equal number of moles of \[{{P}_{4}}\]and \[C{{l}_{2}}\]then obviously at equilibrium\[[{{P}_{4}}]>[C{{l}_{2}}]\].


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